Track / Stage 7 / 126 of 300 #1526 of 2000

Problem 1526

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.2 Prove it Mathematica competitions in Croatia · Croatia

Let $ABC$ be an acute triangle with orthocentre $H$ . The line through the point $A$ perpendicular to $AC$ and the line through the point $B$ perpendicular to $BC$ intersect at $D$ . The circle with centre $C$ which contains $H$ intersects the circumcircle of the triangle $ABC$ at $E$ and $F$ . Prove that $|DE| = |DF| = |AB|$ . (Stipe Vidak)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $\alpha$ , $\beta$ , $\gamma$ be the angles of the triangle $ABC$ and $R$ the radius of its circumcircle. Without loss of generality, let $E$ lie on the arc $BC$ and $F$ on the arc $AC$ .
Since $\angle CAD = \angle CBD = 90^\circ$ , the point $D$ lies on the circumcircle of $ABC$ and $CD$ is the diameter.

Since $|CE| = |CH|$ , $\angle BEC = 180^\circ - \alpha = \angle BHC$ and $BC$ is the common side of $BEC$ and $BHC$ , these triangles are congruent by "S-S-A" theorem (angle $180^\circ - \alpha$ is obtuse). Thus we have $\angle ECB = \angle HCB$ , and the line $BC$ is the axis of symmetry of the isosceles triangle $ECH$ . Therefore $BC \perp HE$ and we conclude that the points $E$ , $H$ and $A$ lie on the same line.

From the right triangle $CDE$ we have
$|DE| = |CD| \cos \angle CDE = |CD| \cos \angle CAE = |CD| \cos(90^\circ - \gamma) = |CD| \sin \gamma.$
By the law of sines we have $2R \sin \gamma = |AB|$ and we get $|DE| = |AB|$ .
Analogously we get $|DF| = |AB|$ .

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